如果另一个值不存在，如何根据一个值从表中选择项目?(雄辩/sql)

I have a table where I store different blocks of texts for my website. I am currently selecting the correct blocks for each page with the slug of the page, and the prefered language. I would like to select the same block of text (with the same title) in the fallback language when the prefered language is not available.

块表

columns:
| id | slug | title | language | content |

entries:
| 1 | home | first | en | the first block |
| 2 | home | first | nl | het eerste blok |
| 3 | home | second | en | the second block |
    --> block "second" not available for the 'nl' language

所需的输出

fallback language = en

selected prefered language = en, output:
| 1 | home | first | en | the first block |
| 3 | home | second | en | the second block |

selected prefered language = nl, output:
| 2 | home | first | nl | het eerste blok |
| 3 | home | second | en | the second block |
     --> select this one, because the 'nl' version is not available

目前我只选择具有所选首选语言的块，因为如果首选语言"块不可用，我不知道如何通过选择后备语言"块来解决此问题.我可以尝试对两种语言运行两个查询，然后以某种方式合并它们，并且仅在此块的标题计数低于 1 时才插入后备语言"块，但这似乎很复杂而且不是很优雅?

$blocks = Block::->where('slug', '=', 'home')
                ->whereIn('language', $selectedLanguage)
                ->get();

select blocks.*
from blocks
left join blocks b1
    on  b1.slug  = blocks.slug
    and b1.title = blocks.title
    and b1.language = 'nl'
    and blocks.language <> 'nl'
where blocks.slug = 'home'
  and blocks.language in ('nl', 'en')
  and b1.id is null

$prefered = 'nl';
$fallback = 'en';

$blocks = AppBlock::where('blocks.slug', '=', 'home')
    ->whereIn('blocks.language', [$prefered, $fallback])
    ->leftJoin('blocks as b1', function($join) {
        $join->on('b1.slug', '=', 'blocks.slug')
             ->on('b1.title', '=', 'blocks.title')
             ->on('b1.language', '=', DB::raw('?'))
             ->on('blocks.language', '<>', DB::raw('?'))
        ;
    })
    ->whereNull('b1.id')
    ->addBinding([$prefered, $prefered], 'join')
    ->select(DB::raw('blocks.*'))
    ->get()
;